Acetylacetone (see Exercise \(45,\) part a), abbreviated acacH, is a bidentate ligand. It loses a proton and coordinates as acac\(^-\) as shown below: Acetylacetone reacts with an ethanol solution containing a salt of europium to give a compound that is 40.1\(\% \mathrm{C}\) and 4.71\(\% \mathrm{H}\) by mass. Combustion of 0.286 \(\mathrm{g}\) of the compound gives 0.112 $\mathrm{g} \mathrm{Eu}_{2} \mathrm{O}_{3} .$ Assuming the compound contains only \(\mathrm{C}, \mathrm{H},\) O, and Eu, determine the formula of the compound formed from the reaction of acetylacetone and the europium salt. (Assume that the compound contains one europium ion.)

Since we know the percent composition of C and H in the compound, we can determine the mass of C and H present in the given 0.286 g sample. Then we can convert this mass into moles. For C: Percent of C = 40.1 % Mass of C = Percent of C × Mass of sample = 0.401 × 0.286 g = 0.114686 g Moles of C = \(\frac{mass\:of\:C}{molecular\:weight\:of\:C}\) = \(\frac{0.114686\: g}{12.01\: g/mol}\) = 0.0095526 mol For H: Percent of H = 4.71 % Mass of H = Percent of H × Mass of sample = 0.0471 × 0.286 g = 0.0134716 g Moles of H = \(\frac{mass\:of\:H}{molecular\:weight\:of\:H}\) = \(\frac{0.0134716\: g}{1.01\: g/mol}\) = 0.0133361 mol

We know that the combustion of 0.286 g of the compound gives 0.112 g of Eu₂O₃. To find the moles of Eu in the compound, we first need to calculate the moles of Eu₂O₃ and then double that value. Molar mass of Eu₂O₃ = (2 × Molar mass of Eu) + (3 × Molar mass of O) = (2 × 151.96 g/mol) + (3 × 16.00 g/mol) = 351.92 g/mol Moles of Eu₂O₃ = \(\frac{mass\:of\:Eu_{2}O_{3}}{molecular\:weight\:of\:Eu_{2}O_{3}}\) = \(\frac{0.112\:g}{351.92\:g/mol}\) = 0.00031807 mol Each mole of Eu₂O₃ contains 2 moles of Eu, so: Moles of Eu = 0.00031807 mol × 2 = 0.00063614 mol

We know that the compound contains only C, H, O, and Eu. We have already determined the moles of C, H, and Eu. We can now find the moles of O by subtracting the sum of the moles of C, H, and Eu from the total moles in the compound. Total moles = \(\frac{mass\:of\:sample}{molecular\:weight\:of\:sample}\) = \(\frac{0.286\:g}{molecular\:weight\:of\:sample}\) To find the total moles, we first need to determine the molecular weight of the compound. We can do this by adding the molecular weights of C, H, O, and Eu to get: Molecular weight of sample = Weight of C + Weight of H + Weight of O + Weight of Eu = 0.114686 g + 0.0134716 g + Weight of O + (0.00063614 mol × 151.96 g/mol) Weight of O = 0.286 g - (0.114686 g + 0.0134716 g + (0.00063614 mol × 151.96 g/mol)) = 0.1346616 g Moles of O = \(\frac{mass\:of\:O}{molecular\:weight\:of\:O}\) = \(\frac{0.1346616\: g}{16.00\: g/mol}\) = 0.00841635 mol

Now that we have the moles of C, H, O, and Eu, we can find the empirical formula of the compound. First, divide the moles of each element by the smallest number of moles: C: \(\frac{0.0095526}{0.00063614}\) = 15.00 H: \(\frac{0.0133361}{0.00063614}\) = 20.98 O: \(\frac{0.00841635}{0.00063614}\) = 13.23 Eu: \(\frac{0.00063614}{0.00063614}\) = 1.00 These values are very close to whole numbers, so we can assume that the empirical formula of the compound is EuC₁₅₀₂₁. Therefore, the formula of the compound formed from the reaction of acetylacetone and the europium salt is EuC₁₅H₂₁O₇.